for k=0.732, the calculator gives the same results as winisd. however, the version of winisd that i have for windows 7 does not allow changing of the end correction to other value, hence the use of the calculator.

with all this considered, i think that a reasonable best guess with this information and targeting a 16-17hz tuning is to use a 28" long slot port with k coming in at somewhere around 1.75.

i will adjust prior posts to reflect this.
So, if folks have more information on this topic, fire away!

Let's look at this a bit more, as I see you have now copied and pasted this in several places.

An overview of the process by which the simulation programs calculate port length is as follows:

Calculate the effective length LEFF of the port based on box volume, box tuning frequency and port cross-sectional area A.

Calculate the actual length of the port LACT using an end correction whose value is proportional to the square root of the port cross-sectional area A. This formula has the form:

LACT = LEFF - k * D

where k is the end correction factor and D is the diameter of the equivalent cylindrical port having cross-sectional area A.

Now, it's been observed that using this equation with k = 0.732 gives incorrect port lengths, and the error results in a port that's usually too long. The website to which you previously linked shows errors between calculated and actual length, but does not establish the why.

For the sake of argument, let's assume your hypothesis is correct, namely that this error is solely due to an error in the end correction factor k. That is, assume there is a value kRIGHT giving the correct length LRIGHT and a value kWRONG giving a wrong length LWRONG. This leads to the following:

LRIGHT = LEFF - kRIGHT * D
LWRONG = LEFF - kWRONG * D

Let's compute the percent error in length as follows:

percent error in length = (LRIGHT - LWRONG) / LRIGHT * 100

In the subtraction of the numerator, LEFF drops out, and the error term in the numerator only depends on the equivalent port diameter D. The result is this:

percent error in length = (kWRONG - kRIGHT) * D / LRIGHT * 100

For a given port area, the numerator does not depend on port length at all. If the port area is constant, the numerator is constant also. That means the percent error in port length would be inversely proportional to the correct port length if this hypothesis were true. Now look at the web site with the data in the table titled "Results, un-damped box". The percent error in port length is almost constant (with actual lengtth / calc. length going from 0.81 to 0.83 as the port length varies by a factor of 3). In the case of 10 cm and 20 cm port lengths, the percent error is the same.

So assuming the data of that web site is correct, the "incorrect end correction" hypothesis fails. So yes, there is an error between computed and actual port lengths, but the assumption that this is due entirely to the end correction factor leads to a contradiction.

For two different port lengths (the right and the wrong one), Fb is not constant. What's constant is LEFF for a given box tuning frequency, box volume and port area. The only variables under consideration here are how L is calculated from LEFF (using different values of k).

But if you use two different values of k to compute the vent length (say, the right one and the wrong one), the actual fb will change as a result of the two different lengths. The desired fb is certainly constant, but you can't just assume the actual fb will not change with a change in vent length.

Edit:

What I'm trying to say is that if you change k in your equation above

Lv = -kDv + cD^2/VbFb^2

namely, take two different values of k and subtract the resulting Lv from each other, the second terms on the right-hand side of the above will not cancel out, because there will be two different fb values involved. This is the actual fb, not the desired one, and the actual fb is a function of k.

we just have a different understanding of what that k is doing. i read the equation as k is the link between a varying port length and a constant box tuning. the intuition seems right to me as well. the more boundaries around the exit of the port, the longer it would seem to the exiting air--synthetically extending the port or something like that.

another oddball that pops up is when a port exits fairly close to a perpendicular boundary, that too reduces effective tuning, but somehow that feels like a different effect than k, almost opposite (flow "jam up" vs. continuation of high velocity).

But there is something else going on. There's a radiation impedance (ratio of complex amplitude of acoustic pressure to volume velocity at the boundary) at each end of the tube. Using a low-frequency approximation, Beranek shows this radiation impedance to be of the mathematical form of an acoustic mass. This is the same exact thing that causes the effective mass of the cone, MMS to be larger than its mechanical mass MMD. This difference between MMS and MMD is the acoustic mass loading of the radiation impedance resulting in an effective mass increase of the cone.

Now, since the two radiation impedance loads on the port are each of the form of an acoustic mass, and the acoustic impedance of the port is also an acoustic mass, the radiation impedances have the effect of making the tube have a larger MA than that given by the formula above, just as the radiation impedance load on the woofer cone makes its mass look bigger. One could think of the larger MA as resulting from an equivalent length LEFF that's longer than its physical length.

is that just semantics or a different idea? it seems that you are saying that a port firing into a smaller space will have greater resistance/mass loading which gives something of a similar effect as a longer port? i'm saying that by extending the boundaries around the exit of the port does the same. what does it matter if we call k end correction factor or mass loading factor?

is that just semantics or a different idea? it seems that you are saying that a port firing into a smaller space will have greater resistance/mass loading which gives something of a similar effect as a longer port?

The "resistance" part is zero. The tube is an acoustic mass, which is a pure acoustic reactance in the calculation. The radiation impedance at each end is also (given a low-frequency approximation) a pure acoustic mass (no resistance, just reactance). So you have three acoustic masses in series in the acoustic equivalent circuit: the tube itself without considering the boundary effects, the radiation impedance on the front side, and the radiation impedance on the back side. And yes, it does give the effect of a longer port. This is covered in many places in the literature.

Quote:

Originally Posted by LTD02

i'm saying that by extending the boundaries around the exit of the port does the same

Huh? The discussion is about whether the discrepancy in calculated and actual length can be explained by only a different end correction factor k. I just demonstrated that if that were true, the percentage error in port length would be inversely proportional to the correct port length. It isn't. In fact, it's darned near constant, which shows that something else is going on.

Quote:

Originally Posted by LTD02

what does it matter if we call k end correction factor or mass loading factor?

It doesn't. It was just an attempt to clear up some confusion in communication.

I was asking more in the context of the effective tuning frequency. it seems that the impedance mismatch between the air slug in the port and the external air may be creating a condition where the effective tuning frequency of the port is much lower than predicted by the equation above.

Trying this again
Throwing this out there but if u have a port dia. Of 6in then then end of your port should be about 6in away or more from the back of your enclosure or it would add and unknown length to the port?

"Trying this again
Throwing this out there but if u have a port dia. Of 6in then then end of your port should be about 6in away or more from the back of your enclosure or it would add and unknown length to the port?"

yep. that is as I understand it. once you get less than 1 port width to a boundary, turbulence builds up and the flow is restricted. that creates something of the same effect as increasing the length of the port, but it can also cause compression if not done properly.

"Trying this again
Throwing this out there but if u have a port dia. Of 6in then then end of your port should be about 6in away or more from the back of your enclosure or it would add and unknown length to the port?"

yep. that is as I understand it. once you get less than 1 port width to a boundary, turbulence builds up and the flow is restricted. that creates something of the same effect as increasing the length of the port, but it can also cause compression if not done properly.

So then in the annihilator ill only have 4in est. to the back of the enclosure i would need more depth or turn the ports up?

but what about the impedance mismatch that andy was talking about noah?

I didn't say anything at all about impedance matching or mismatch. The purpose of impedance matching is to maximize delivered power, but as Noah correctly points out, that doesn't apply in the case of low-efficiency systems.